jmc

Using the identity ${\log }_{10}(x^2)=2{\log }_{10}x,$ the first equation simplifies to $$({\log }_{10}x{)}^2-2{\log }_{10}x=48.$$Subtracting $48$ from both sides gives a quadratic equation in ${\log }_{10}x,$ which factors as $$({\log }_{10}x-8)({\log }_{10}x+6)=0.$$Since $x<1,$ we have ${\log }_{10}x<0,$ so we must choose the negative root, ${\log }_{10}x=-6.$ Then using the identity ${\log }_{10}(x^3)=3{\log }_{10}x$ gives the answer: $$\begin{array}{rl}({\log }_{10}x{)}^3-{\log }_{10}{x}^3& =({\log }_{10}x{)}^3-3{\log }_{10}x\\ & =(-6{)}^3-3(-6)\\ & =-216+18\\ & =\boxed{-198}.\end{array}$$